n (or m) must equal 1 but r must also equal 1 because you only sum over two terms. But r obviously cannot be 1 since it is a + 1 on the right side. I am thinking that there some other forms of Vandermonde's identity which it may be a special case of but not this one? This form does not even make sense when r is greater than n and we need r to be general and n to be 1 I think...

You possibly end up with [tex] \binom{m+1}{a+1} = \sum_{k=0}^{a+1}
\binom{m}{a+1-k} * \binom{1}{k} [/tex]

I am not sure but , I guess [tex] \binom{1}{k} =0[/tex] if k>1 by the definition of combination
Because the number of ways picking e.g. 3 elements from a set with 1 element must be zero.
So then you can get what you want only summing two terms as you said
n